Calculus I, II, II, and differential equations

College Park Tutors Blog

College Park Tutors Blog

Mar
22,
2015

Or more specifically, a second-order *linear* homogeneous differential equation with complex roots. Yeesh, its always a mouthful with diff eq. Oh and, we'll throw in an initial condition just for sharks and goggles. The problem goes like this:

Find a real-valued solution to the initial value problem \(y''+4y=0\), with \(y(0)=0\) and \(y'(0)=1\). Your solution must be real-valued or you will not receive full credit!

If you'll recall, the steps for solving a second-order homogeneous diff. eq. are as follows:

- Write down the characteristic equation.
- Find the roots of the characteristic equation.
- Use the roots to write down the two exponential basis solutions.
- Create a general solution using a linear combination of the two basis solutions.

For step 1, we simply take our differential equation and replace \(y''\) with \(r^2\), \(y'\) with \(r\), and \(y\) with 1. Easy enough:

For step 2, we solve this quadratic equation to get two roots. The roots are going to be complex numbers, but that's ok:

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Mar
27,
2015

If you're reading this post, you've probably arrived at the point in your calculus course where they try to teach you how to do integration by partial fraction expansion. And if you're anything like I was when I first learned calculus, you're probably scratching your head and going **whadafuhhhh**? **What is the point of all this**?

Here's what you need to know. *Partial fraction expansion is not an integration technique*. It's an algebraic technique. That being said, it's useful for making certain algebraic expressions (i.e. rational expressions) easier to integrate by breaking them into smaller, simpler chunks. Sound familiar? Its that same mode of thinking that all your other calculus classes are pushing at - taking a difficult problem, decomposing it into several easier problems, and then putting the pieces back together.

Got it? Let's get down to business. Here's our problem:

Rewrite the following expression as the sum of three partial fraction terms: \[\frac{1}{(x-1)(x^2+x+1)}\]

Just to quickly review the step...